【发布时间】：2014-03-30 07:27:31
【问题描述】：
我最近有一个朋友专门研究梯形逻辑而不是网络编程，来找我寻求她雇主的项目帮助。虽然我使用更传统的编码语言，但我自己远不是 jquery 和 php 方面的专家。我们遇到的问题是通过 XMLHttpRequest 将 jquery / html 表单插入父页面的 php 页面没有从父页面执行其“发布”操作。使这个问题变得更加困难的是，当页面在父页面之外（直接加载到浏览器中）自行运行时，它会很好地执行其“发布”操作。在这一点上，我已经进行了许多小时的搜索和反复试验，但我很困惑，现在向您寻求帮助。以下是相关的代码位。您可以提供的任何帮助将不胜感激，因为在通过 XMLHttpRequest 插入表单时执行提交表单时，我们尝试过的任何方法似乎都不起作用。
父页面的Javascript代码插入外部表单：
function showUser(str)
{
if (str=="")
{
document.getElementById("insertUserHere").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp2=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp2=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp2.onreadystatechange=function()
{
if (xmlhttp2.readyState==4 && xmlhttp.status==200)
{
document.getElementById("insertUserHere").innerHTML=xmlhttp2.responseText;
}
}
xmlhttp2.open("GET","ajax-userForm.php?q="+str,true);
xmlhttp2.send();
}
由 xhmlhttprequest (ajax-userForm.php) 插入的外部 PHP 页面的代码：
<script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
<script src="http://code.jquery.com/jquery-migrate-1.2.1.min.js"></script>
<script type="text/javascript">
// JQUERY: Plugin "autoSubmit"
(function($) {
$.fn.autoSubmit = function(options) {
return $.each(this, function() {
// VARIABLES: Input-specific
var input = $(this);
var column = input.attr('name');
// VARIABLES: Form-specific
var form = input.parents('form');
var method = form.attr('method');
var action = form.attr('action');
// VARIABLES: Where to update in database
var where_val = form.find('#where').val();
var where_col = form.find('#where').attr('name');
// ONBLUR: Dynamic value send through Ajax
input.bind('blur', function(event) {
// Get latest value
var value = input.val();
if (input.attr('type') == "checkbox")
{
if (input.attr('checked') )
{
value = 1;
}
else
{
value = 0;
}
}
// AJAX: Send values
$.ajax({
url: action,
type: method,
data: {
val: value,
col: column,
w_col: where_col,
w_val: where_val
},
cache: false,
timeout: 10000,
success: function(data) {
// Alert if update failed
if (data) {
alert(data);
}
// Load output into a P
else {
$('#notice').text('Updated');
$('#notice').fadeOut().fadeIn();
}
}
});
// Prevent normal submission of form
return false;
})
});
}
})(jQuery);
// JQUERY: Run .autoSubmit() on all INPUT fields within form
$(function(){
$('#ajax-userForm INPUT').autoSubmit();
});
</script>
<!-- STYLE -->
<style type="text/css">
INPUT { margin-right: 1em }
</style>
</head>
<body>
<!-- CONTENT -->
<?php
$q = intval($_GET['q']);
/*
* DATABASE CONNECTION
*/
// DATABASE: Connection variables
$db_host = "localhost";
$db_name = "DBNAME";
$db_username = "root";
$db_password = "DBPWD";
// DATABASE: Try to connect
if (!$db_connect = mysql_connect($db_host, $db_username, $db_password))
die('Unable to connect to MySQL from ajax-form.');
if (!$db_select = mysql_select_db($db_name, $db_connect))
die('Unable to select database');
/*
* DATABASE QUERY
*/
// DATABASE: Get current row
//$result = mysql_query("SELECT * FROM user");
$result = mysql_query("SELECT * FROM user where Project_ID = '".$q."' ");
$row = mysql_fetch_assoc($result);
?>
<form id="ajax-userForm" class="autosubmit" method="post" action="ajax-updateUser.php">
<fieldset>
<legend>Update user information</legend>
<label>First Name:</label>
<input name="FirstName" value="<?php echo $row['FirstName'] ?>" />
<label>Last Name:</label>
<input name="LastName" value="<?php echo $row['LastName'] ?>" />
<label>Hometown</label>
<input name="Hometown" value="<?php echo $row['Hometown'] ?>" />
<label>Married</label>
<input type = "checkbox" id = "chkMarried" name="Married" <?php echo $row['Married'] == 1 ? 'checked':'unchecked' ?>/>
<label>Employed</label>
<input type = "checkbox" id = "chkEmployed" name="Employed" <?php echo $row['Employed'] == 1 ? 'checked':'unchecked' ?> />
<input id="where" type="hidden" name="Project_ID" value="<?php echo $row['Project_ID'] ?>" />
</fieldset>
</form>
<p id="notice"></p>
</body>
</html>
页面代码 (ajax-updateUser.php) 由上述代码 (ajax-userForm.php) 中的“发布”操作调用：
/*
* DATABASE CONNECTION
*/
// DATABASE: Connection variables
$db_host = "localhost";
$db_name = "DBNAME";
$db_username = "root";
$db_password = "DBPWD";
// DATABASE: Try to connect
if (!$db_connect = mysql_connect($db_host, $db_username, $db_password))
die('Unable to connect to MySQL from ajax-update.');
if (!$db_select = mysql_select_db($db_name, $db_connect))
die('Unable to select database');
$message = "Connection Successful";
//echo "<script type='text/javascript'>alert('$message');</script>";
// DATABASE: Clean data before use
function clean($value)
{
return mysql_real_escape_string($value);
}
/*
* FORM PARSING
*/
// FORM: Variables were posted
if (count($_POST) > 0)
{
$message = count($_POST);
//echo "<script type='text/javascript'>alert('$message');</script>";
// Prepare form variables for database
foreach($_POST as $column => $value)
${$column} = clean($value);
// Perform MySQL UPDATE
$result = mysql_query("UPDATE user SET ".$col."='".$val."'
WHERE ".$w_col."='".$w_val."'")
or die ('Unable to update row.');
}
else
{
$message = "Nothing in Post";
echo "<script type='text/javascript'>alert('$message');</script>";
}
?>
【问题讨论】：
-
旁注：您确实意识到您的 PHP/SQL 有问题。注意到 SO 的语法高亮是如何表现的吗？真的不多，但我敢打赌这只是一个错字，而不是你的工作代码;-)
-
是的，我认为它只是在我在 SO 上插入代码时发生的。当我使用虚拟数据库密码发布我的版本时，我也忘记关闭报价。
标签： javascript php jquery ajax xmlhttprequest